Problem: Simplify and expand the following expression: $ \dfrac{q - 2}{3q - 8}+\dfrac{q}{4q + 6} $
In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(3q - 8)(4q + 6)$ Multiply the first term by $\dfrac{4q + 6}{4q + 6}$ $ \begin{align*} \dfrac{q - 2}{3q - 8} \times \dfrac{4q + 6}{4q + 6} & = \dfrac{(q - 2)(4q + 6)}{(3q - 8)(4q + 6)} \\ & = \dfrac{4q^2 - 2q - 12}{(3q - 8)(4q + 6)}\end{align*} $ Multiply the second term by $\dfrac{3q - 8}{3q - 8}$ $ \begin{align*} \dfrac{q}{4q + 6} \times \dfrac{3q - 8}{3q - 8} & = \dfrac{(q)(3q - 8)}{(4q + 6)(3q - 8)} \\ & = \dfrac{3q^2 - 8q}{(4q + 6)(3q - 8)}\end{align*} $ Now we have: $ = \dfrac{4q^2 - 2q - 12}{(3q - 8)(4q + 6)} + \dfrac{3q^2 - 8q}{(4q + 6)(3q - 8)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{4q^2 - 2q - 12 + 3q^2 - 8q}{(3q - 8)(4q + 6)} $ $ = \dfrac{7q^2 - 10q - 12}{(3q - 8)(4q + 6)}$ Expand the denominator: $ = \dfrac{7q^2 - 10q - 12}{12q^2 - 14q - 48}$